{"id":1153,"date":"2021-06-30T17:02:00","date_gmt":"2021-06-30T17:02:00","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/problem-set-the-definite-integral\/"},"modified":"2021-12-08T23:58:17","modified_gmt":"2021-12-08T23:58:17","slug":"problem-set-the-definite-integral","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/problem-set-the-definite-integral\/","title":{"raw":"Problem Set: The Definite Integral","rendered":"Problem Set: The Definite Integral"},"content":{"raw":"

In the following exercises, express the limits as integrals.<\/p>\r\n\r\n

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1.\u00a0<\/strong>[latex]\\underset{n\\to \\infty }{\\lim}\\underset{i=1}{\\overset{n}{\\Sigma}}(x_i^*) \\Delta x[\/latex] over [latex][1,3][\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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2.\u00a0<\/strong>[latex]\\underset{n\\to \\infty }{\\lim}\\underset{i=1}{\\overset{n}{\\Sigma}}(5(x_i^*)^2-3(x_i^*)^3) \\Delta x[\/latex] over [latex][0,2][\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572456371\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572456371\"]\r\n

[latex]\\displaystyle\\int_0^2 (5x^2-3x^3) dx[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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3.\u00a0<\/strong>[latex]\\underset{n\\to \\infty }{\\lim}\\underset{i=1}{\\overset{n}{\\Sigma}} \\sin^2 (2\\pi x_i^*) \\Delta x[\/latex] over [latex][0,1][\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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4.\u00a0<\/strong>[latex]\\underset{n\\to \\infty }{\\lim}\\underset{i=1}{\\overset{n}{\\Sigma}} \\cos^2 (2\\pi x_i^*) \\Delta x[\/latex] over [latex][0,1][\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572218601\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572218601\"]\r\n

[latex]\\displaystyle\\int_0^1 \\cos^2 (2\\pi x) dx[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

In the following exercises, given [latex]L_n[\/latex]\u00a0or [latex]R_n[\/latex]\u00a0as indicated, express their limits as [latex]n\\to \\infty [\/latex] as definite integrals, identifying the correct intervals.<\/p>\r\n\r\n

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5.\u00a0<\/strong>[latex]L_n=\\frac{1}{n}\\underset{i=1}{\\overset{n}{\\Sigma}}\\frac{i-1}{n}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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6.\u00a0<\/strong>[latex]R_n=\\frac{1}{n}\\underset{i=1}{\\overset{n}{\\Sigma}}\\frac{i}{n}[\/latex]<\/p>\r\n[reveal-answer q=\"776574\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"776574\"][latex]\\displaystyle\\int_0^1 x dx[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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7.\u00a0<\/strong>[latex]L_n=\\frac{2}{n}\\underset{i=1}{\\overset{n}{\\Sigma}}(1+2\\frac{i-1}{n})[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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8.\u00a0<\/strong>[latex]R_n=\\frac{3}{n}\\underset{i=1}{\\overset{n}{\\Sigma}}(3+3\\frac{i}{n})[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170571710642\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571710642\"]\r\n

[latex]\\displaystyle\\int_3^6 x dx[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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9.\u00a0<\/strong>[latex]L_n=\\frac{2\\pi }{n}\\underset{i=1}{\\overset{n}{\\Sigma}}2\\pi \\frac{i-1}{n} \\cos (2\\pi \\frac{i-1}{n})[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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10.\u00a0<\/strong>[latex]R_n=\\frac{1}{n}\\underset{i=1}{\\overset{n}{\\Sigma}}(1+\\frac{i}{n})\\log((1+\\frac{i}{n})^2)[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572168726\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572168726\"][latex]\\displaystyle\\int_1^2 x\\log(x^2) dx[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

In the following exercises (11-16), evaluate the integrals of the functions graphed using the formulas for areas of triangles and circles, and subtracting the areas below the [latex]x[\/latex]-axis.<\/p>\r\n\r\n

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11.\u00a0<\/strong>\"A<\/span><\/div>\r\n<\/div>\r\n
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\r\n\r\n12.\u00a0<\/strong>\"A<\/span>\r\n
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[latex]1+2 \\cdot 2+3 \\cdot 3=14[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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13.\u00a0<\/strong>\"A<\/span><\/div>\r\n<\/div>\r\n
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14.\u00a0<\/strong>\"A<\/span>\r\n[reveal-answer q=\"fs-id1170571689807\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571689807\"]\r\n

[latex]1-4+9=6[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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15.\u00a0<\/strong>\"A<\/span><\/div>\r\n<\/div>\r\n
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\r\n\r\n16.\u00a0<\/strong>\"A<\/span>\r\n
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[latex]1-2\\pi +9=10-2\\pi [\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

In the following exercises (17-24), evaluate the integral using area formulas.<\/p>\r\n\r\n

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17.\u00a0<\/strong>[latex]\\displaystyle\\int_0^3 (3-x) dx[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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18.\u00a0<\/strong>[latex]\\displaystyle\\int_2^3 (3-x) dx[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170571777869\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571777869\"]\r\n

The integral is the area of the triangle, [latex]\\frac{1}{2}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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19.\u00a0<\/strong>[latex]\\displaystyle\\int_{-3}^3 (3-|x|) dx[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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20.\u00a0<\/strong>[latex]\\displaystyle\\int_0^6 (3-|x-3|) dx[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572569972\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572569972\"]\r\n

The integral is the area of the triangle, 9.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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21.\u00a0<\/strong>[latex]\\displaystyle\\int_{-2}^2 \\sqrt{4-x^2} dx[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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22.\u00a0<\/strong>[latex]\\displaystyle\\int_1^5 \\sqrt{4-(x-3)^2} dx[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572310044\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572310044\"]The integral is the area [latex]\\frac{1}{2}\\pi r^2=2\\pi[\/latex].[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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23.\u00a0<\/strong>[latex]\\displaystyle\\int_0^{12} \\sqrt{36-(x-6)^2} dx[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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24.\u00a0<\/strong>[latex]\\displaystyle\\int_{-2}^3 (3-|x|) dx[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572347002\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572347002\"]\r\n

The integral is the area of the \u201cbig\u201d triangle minus the \u201cmissing\u201d triangle, [latex]9-\\frac{1}{2}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

In the following exercises (25-28), use averages of values at the left ([latex]L[\/latex]) and right ([latex]R[\/latex]) endpoints to compute the integrals of the piecewise linear functions with graphs that pass through the given list of points over the indicated intervals.<\/p>\r\n\r\n

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\r\n\r\n25.\u00a0<\/strong>[latex]\\{(0,0),(2,1),(4,3),(5,0),(6,0),(8,3)\\}[\/latex] over [latex][0,8][\/latex]\r\n\r\n<\/div>\r\n<\/div>\r\n
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26.\u00a0<\/strong>[latex]\\{(0,2),(1,0),(3,5),(5,5),(6,2),(8,0)\\}[\/latex] over [latex][0,8][\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572379158\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572379158\"]\r\n

[latex]L=2+0+10+5+4=21,R=0+10+10+2+0=22, \\, \\frac{L+R}{2}=21.5[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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27.\u00a0<\/strong>[latex]\\{(-4,-4),(-2,0),(0,-2),(3,3),(4,3)\\}[\/latex] over [latex][-4,4][\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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28.\u00a0<\/strong>[latex]\\{(-4,0),(-2,2),(0,0),(1,2),(3,2),(4,0)\\}[\/latex] over [latex][-4,4][\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572504514\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572504514\"]\r\n

[latex]L=0+4+0+4+2=10,R=4+0+2+4+0=10, \\, \\frac{L+R}{2}=10[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

Suppose that [latex]\\displaystyle\\int_0^4 f(x) dx=5[\/latex] and [latex]\\displaystyle\\int_0^2 f(x) dx=-3[\/latex], and [latex]\\displaystyle\\int_0^4 g(x) dx=-1[\/latex] and [latex]\\displaystyle\\int_0^2 g(x) dx=2[\/latex]. In the following exercises (29-34), compute the integrals.<\/p>\r\n\r\n

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29.\u00a0<\/strong>[latex]\\displaystyle\\int_0^4 (f(x)+g(x)) dx[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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30.\u00a0<\/strong>[latex]\\displaystyle\\int_2^4 (f(x)+g(x)) dx[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572379216\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572379216\"]\r\n

[latex]\\displaystyle\\int_2^4 f(x) dx + \\displaystyle\\int_2^4 g(x) dx=8-3=5[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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31.\u00a0<\/strong>[latex]\\displaystyle\\int_0^2 (f(x)-g(x)) dx[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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32.\u00a0<\/strong>[latex]\\displaystyle\\int_2^4 (f(x)-g(x)) dx[\/latex]<\/p>\r\n[reveal-answer q=\"336759\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"336759\"][latex]\\displaystyle\\int_2^4 f(x) dx - \\displaystyle\\int_2^4 g(x) dx=8+3=11[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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33.\u00a0<\/strong>[latex]\\displaystyle\\int_0^2 (3f(x)-4g(x)) dx[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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34.\u00a0<\/strong>[latex]\\displaystyle\\int_2^4 (4f(x)-3g(x)) dx[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170571547405\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571547405\"]\r\n

[latex]4 \\displaystyle\\int_2^4 f(x) dx - 3 \\displaystyle\\int_2^4 g(x) dx = 32+9=41[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

In the following exercises (35-38), use the identity [latex]\\displaystyle\\int_{\u2212A}^A f(x) dx = \\displaystyle\\int_{\u2212A}^0 f(x) dx + \\displaystyle\\int_0^A f(x) dx[\/latex] to compute the integrals.<\/p>\r\n\r\n

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\r\n\r\n35.\u00a0<\/strong>[latex]\\displaystyle\\int_{\u2212\\pi}^{\\pi} \\frac{\\sin t}{1+t^2} dt[\/latex] (Hint<\/em>: [latex]\\sin(\u2212t)=\u2212\\sin (t)[\/latex])\r\n\r\n<\/div>\r\n<\/div>\r\n
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36. <\/strong>[latex]\\displaystyle\\int_{\u2212\\sqrt{\\pi}}^{\\sqrt{\\pi}} \\frac{t}{1+ \\cos t} dt[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572351512\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572351512\"]\r\n

The integrand is odd; the integral is zero.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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37.\u00a0<\/strong>[latex]\\displaystyle\\int_1^3 (2-x) dx[\/latex] (Hint:<\/em> Look at the graph of [latex]f[\/latex].)<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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38.\u00a0<\/strong>[latex]{\\displaystyle\\int }_{2}^{4}{(x-3)}^{3}dx[\/latex] (Hint:<\/em> Look at the graph of [latex]f[\/latex].)<\/p>\r\n[reveal-answer q=\"fs-id1170571638086\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571638086\"]\r\n

The integrand is antisymmetric with respect to [latex]x=3[\/latex]. The integral is zero.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

In the following exercises (39-44), given that [latex]\\displaystyle\\int_0^1 x dx = \\frac{1}{2}, \\, \\displaystyle\\int_0^1 x^2 dx = \\frac{1}{3}[\/latex], and [latex]\\displaystyle\\int_0^1 x^3 dx = \\frac{1}{4}[\/latex], compute the integrals.<\/p>\r\n\r\n

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39.\u00a0<\/strong>[latex]\\displaystyle\\int_0^1 (1+x+x^2+x^3) dx[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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40.\u00a0<\/strong>[latex]\\displaystyle\\int_0^1 (1-x+x^2-x^3) dx[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572448502\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572448502\"]\r\n

[latex]1-\\frac{1}{2}+\\frac{1}{3}-\\frac{1}{4}=\\frac{7}{12}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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41.\u00a0<\/strong>[latex]\\displaystyle\\int_0^1 (1-x)^2 dx[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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42.\u00a0<\/strong>[latex]\\displaystyle\\int_0^1 (1-2x)^3 dx[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572306426\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572306426\"]\r\n

[latex]\\displaystyle\\int_0^1 (1-6x+12x^2-8x^3) dx = (x-3x^{2}+4x^{3}-2x^{4})=(1-3+4-2)=0[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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43.\u00a0<\/strong>[latex]\\displaystyle\\int_0^1 (6x-\\frac{4}{3}x^2) dx[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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44.\u00a0<\/strong>[latex]\\displaystyle\\int_0^1 (7-5x^3) dx[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572558087\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572558087\"]\r\n

[latex]7-\\frac{5}{4}=\\frac{23}{4}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

In the following exercises (45-50), use the comparison theorem.<\/p>\r\n\r\n

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45.\u00a0<\/strong>Show that [latex]\\displaystyle\\int_0^3 (x^2-6x+9) dx \\ge 0[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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46.\u00a0<\/strong>Show that [latex]\\displaystyle\\int_{-2}^3 (x-3)(x+2) dx \\le 0[\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1170572307627\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572307627\"]\r\n

The integrand is negative over [latex][-2,3][\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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47.\u00a0<\/strong>Show that [latex]\\displaystyle\\int_0^1 \\sqrt{1+x^3} dx \\le \\displaystyle\\int_0^1 \\sqrt{1+x^2} dx[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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48.\u00a0<\/strong>Show that [latex]\\displaystyle\\int_1^2 \\sqrt{1+x} dx \\le \\displaystyle\\int_1^2 \\sqrt{1+x^2} dx[\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1170572624744\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572624744\"]\r\n

[latex]x \\le x^2[\/latex] over [latex][1,2][\/latex], so [latex]\\sqrt{1+x} \\le \\sqrt{1+x^2}[\/latex] over [latex][1,2][\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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49.\u00a0<\/strong>Show that [latex]\\displaystyle\\int_0^{\\pi\/2} \\sin [latex]t[\/latex] dt \\ge \\frac{\\pi}{4}[\/latex]. (Hint<\/em>: [latex]\\sin [latex]t[\/latex] \\ge \\frac{2t}{\\pi}[\/latex] over [latex][0,\\frac{\\pi}{2}][\/latex])<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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50.\u00a0<\/strong>Show that [latex]\\displaystyle\\int_{\u2212\\pi\/4}^{\\pi\/4} \\cos [latex]t[\/latex] dt \\ge \\pi \\sqrt{2}\/4[\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1170571712451\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571712451\"]\r\n

[latex] \\cos (t) \\ge \\frac{\\sqrt{2}}{2}[\/latex]. Multiply by the length of the interval to get the inequality.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

In the following exercises (51-56), find the average value [latex]f_{\\text{ave}}[\/latex]\u00a0of [latex]f[\/latex] between [latex]a[\/latex] and [latex]b[\/latex], and find a point [latex]c[\/latex], where [latex]f(c)=f_{\\text{ave}}[\/latex].<\/p>\r\n\r\n

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51.\u00a0<\/strong>[latex]f(x)=x^2, \\, a=-1, \\, b=1[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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52.\u00a0<\/strong>[latex]f(x)=x^5, \\, a=-1, \\, b=1[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170571542856\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571542856\"]\r\n

[latex]f_{\\text{ave}}=0; \\, c=0[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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53.\u00a0<\/strong>[latex]f(x)=\\sqrt{4-x^2}, \\, a=0, \\, b=2[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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54.\u00a0<\/strong>[latex]f(x)=(3-|x|), \\, a=-3, \\, b=3[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170571698168\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571698168\"]\r\n

[latex]\\frac{3}{2}[\/latex] when [latex]c= \\pm \\frac{3}{2}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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55.\u00a0<\/strong>[latex]f(x)= \\sin x, \\, a=0, \\, b=2\\pi [\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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56.\u00a0<\/strong>[latex]f(x)= \\cos x, \\, a=0, \\, b=2\\pi [\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170572309610\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572309610\"]\r\n

[latex]f_{\\text{ave}}=0; \\, c=\\frac{\\pi}{2},\\frac{3\\pi}{2}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

In the following exercises, approximate the average value using Riemann sums [latex]L_{100}[\/latex]\u00a0and [latex]R_{100}[\/latex]. How does your answer compare with the exact given answer?<\/p>\r\n\r\n

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57.\u00a0[T]<\/strong> [latex]y=\\ln (x)[\/latex] over the interval [latex][1,4][\/latex]; the exact solution is [latex]\\dfrac{\\ln (256)}{3}-1[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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58. [T]\u00a0<\/strong>[latex]y=e^{x\/2}[\/latex] over the interval [latex][0,1][\/latex]; the exact solution is [latex]2(\\sqrt{e}-1)[\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1170572554275\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572554275\"]\r\n

[latex]L_{100}=1.294, \\, R_{100}=1.301[\/latex]; the exact average is between these values.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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59. [T]\u00a0<\/strong>[latex]y= \\tan x[\/latex] over the interval [latex][0,\\frac{\\pi}{4}][\/latex]; the exact solution is [latex]\\dfrac{2\\ln (2)}{\\pi}[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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60. [T]\u00a0<\/strong>[latex]y=\\dfrac{x+1}{\\sqrt{4-x^2}}[\/latex] over the interval [latex][-1,1][\/latex]; the exact solution is [latex]\\frac{\\pi }{6}[\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1170571637418\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571637418\"]\r\n

[latex]L_{100} \\times (\\frac{1}{2})=0.5178, \\, R_{100} \\times (\\frac{1}{2})=0.5294[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

In the following exercises, compute the average value using the left Riemann sums [latex]L_N[\/latex]\u00a0for [latex]N=1,10,100[\/latex]. How does the accuracy compare with the given exact value?<\/p>\r\n\r\n

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61. [T]\u00a0<\/strong>[latex]y=x^2-4[\/latex] over the interval [latex][0,2][\/latex]; the exact solution is [latex]-\\frac{8}{3}[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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62. [T]\u00a0<\/strong>[latex]y=xe^{x^2}[\/latex] over the interval [latex][0,2][\/latex]; the exact solution is [latex]\\frac{1}{4}(e^4-1)[\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1170572373490\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572373490\"]\r\n

[latex]L_1=0, \\, L_{10} \\times (\\frac{1}{2})=8.743493, \\, L_{100} \\times (\\frac{1}{2})=12.861728[\/latex]. The exact answer [latex]\\approx 26.799[\/latex], so [latex]L_{100}[\/latex]\u00a0is not accurate.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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63. [T]\u00a0<\/strong>[latex]y=\\left(\\frac{1}{2}\\right)^x[\/latex] over the interval [latex][0,4][\/latex]; the exact solution is [latex]\\dfrac{15}{64\\ln (2)}[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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64. [T]\u00a0<\/strong>[latex]y=x \\sin (x^2)[\/latex] over the interval [latex][\u2212\\pi ,0][\/latex]; the exact solution is [latex]\\dfrac{\\cos (\\pi^2)-1}{2\\pi}[\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1170572172959\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572172959\"]\r\n

[latex]L_1 \\times (\\frac{1}{\\pi})=1.352, \\, L_{10} \\times (\\frac{1}{\\pi})=-0.1837, \\, L_{100} \\times (\\frac{1}{\\pi})=-0.2956[\/latex]. The exact answer [latex]\\approx -0.303[\/latex], so [latex]L_{100}[\/latex]\u00a0is not accurate to first decimal.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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65.\u00a0<\/strong>Suppose that [latex]A=\\displaystyle\\int_0^{2\\pi} \\sin^2 [latex]t[\/latex] dt[\/latex] and [latex]B=\\displaystyle\\int_0^{2\\pi} \\cos^2 [latex]t[\/latex] dt[\/latex]. Show that [latex]A+B=2\\pi [\/latex] and [latex]A=B[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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66.\u00a0<\/strong>Suppose that [latex]A=\\displaystyle\\int_{\u2212\\pi\/4}^{\\pi\/4} \\sec^2 [latex]t[\/latex] dt = \\pi [\/latex] and [latex]B=\\displaystyle\\int_{\u2212\\pi\/4}^{\\pi\/4} \\tan^2 [latex]t[\/latex] dt[\/latex]. Show that [latex]B-A=\\frac{\\pi }{2}[\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1170571614590\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571614590\"]\r\n

Use [latex]\\tan^2 \\theta +1= \\sec^2 \\theta[\/latex]. Then, [latex]B-A=\\displaystyle\\int_{\u2212\\pi\/4}^{\\pi\/4} 1 dx = \\frac{\\pi}{2}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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67.\u00a0<\/strong>Show that the average value of [latex]\\sin^2 t[\/latex] over [latex][0,2\\pi][\/latex] is equal to [latex]\\frac{1}{2}[\/latex]. Without further calculation, determine whether the average value of [latex]\\sin^2 t[\/latex] over [latex][0,\\pi][\/latex] is also equal to [latex]\\frac{1}{2}[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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68.\u00a0<\/strong>Show that the average value of [latex]\\cos^2 t[\/latex] over [latex][0,2\\pi][\/latex] is equal to [latex]1\/2[\/latex]. Without further calculation, determine whether the average value of [latex]\\cos^2 (t)[\/latex] over [latex][0,\\pi][\/latex] is also equal to [latex]1\/2[\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1170572372085\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572372085\"]\r\n

[latex]\\displaystyle\\int_0^{2\\pi} \\cos^2 [latex]t[\/latex] dt = \\pi[\/latex], so divide by the length [latex]2\\pi[\/latex] of the interval. [latex]\\cos^2 t[\/latex] has period [latex]\\pi[\/latex], so yes, it is true.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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69.\u00a0<\/strong>Explain why the graphs of a quadratic function (parabola) [latex]p(x)[\/latex] and a linear function [latex]\\ell (x)[\/latex] can intersect in at most two points. Suppose that [latex]p(a)=\\ell (a)[\/latex] and [latex]p(b)=\\ell (b)[\/latex], and that [latex]\\displaystyle\\int_a^b p(t) dt > \\displaystyle\\int_a^b \\ell (t) dt[\/latex]. Explain why [latex]\\displaystyle\\int_c^d p(t) > \\displaystyle\\int_c^d \\ell (t) dt[\/latex] whenever [latex]a \\le c < d \\le b[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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70.\u00a0<\/strong>Suppose that parabola [latex]p(x)=ax^2+bx+c[\/latex] opens downward [latex](a<0)[\/latex] and has a vertex of [latex]y=\\frac{\u2212b}{2a}>0[\/latex]. For which interval [latex][A,B][\/latex] is [latex]\\displaystyle\\int_A^B (ax^2+bx+c) dx[\/latex] as large as possible?<\/p>\r\n[reveal-answer q=\"fs-id1170572274654\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572274654\"]\r\n

The integral is maximized when one uses the largest interval on which [latex]p[\/latex] is nonnegative. Thus, [latex]A=\\frac{\u2212b-\\sqrt{b^2-4ac}}{2a}[\/latex] and [latex]B=\\frac{\u2212b+\\sqrt{b^2-4ac}}{2a}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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71.\u00a0<\/strong>Suppose [latex][a,b][\/latex] can be subdivided into subintervals [latex]a=a_0<a_1<a_2< \\cdots <a_N=b[\/latex] such that either [latex]f\\ge 0[\/latex] over [latex][a_{i-1},a_i][\/latex] or [latex]f\\le 0[\/latex] over [latex][a_{i-1},a_i][\/latex]. Set [latex]A_i=\\displaystyle\\int_{a_{i-1}}^{a_i} f(t) dt[\/latex].<\/p>\r\n\r\n

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  1. Explain why [latex]\\displaystyle\\int_a^b f(t) dt = A_1+A_2+ \\cdots +A_N[\/latex].<\/li>\r\n \t
  2. Then, explain why [latex]|\\displaystyle\\int_a^b f(t) dt| \\le \\displaystyle\\int_a^b |f(t)| dt[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n
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    72.\u00a0<\/strong>Suppose [latex]f[\/latex] and [latex]g[\/latex] are continuous functions such that [latex]\\displaystyle\\int_c^d f(t) dt \\le \\displaystyle\\int_c^d g(t) dt[\/latex] for every subinterval [latex][c,d][\/latex] of [latex][a,b][\/latex]. Explain why [latex]f(x)\\le g(x)[\/latex] for all values of [latex]x[\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1170571596326\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571596326\"]\r\n

    If [latex]f(t_0)>g(t_0)[\/latex] for some [latex]t_0 \\in [a,b][\/latex], then since [latex]f-g[\/latex] is continuous, there is an interval containing [latex]t_0[\/latex] such that [latex]f(t)>g(t)[\/latex] over the interval [latex][c,d][\/latex], and then [latex]\\displaystyle\\int_c^d f(t) dt>\\displaystyle\\int_c^d g(t) dt[\/latex] over this interval.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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    73.\u00a0<\/strong>Suppose the average value of [latex]f[\/latex] over [latex][a,b][\/latex] is 1 and the average value of [latex]f[\/latex] over [latex][b,c][\/latex] is 1 where [latex]a<c<b[\/latex]. Show that the average value of [latex]f[\/latex] over [latex][a,c][\/latex] is also 1.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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    74.\u00a0<\/strong>Suppose that [latex][a,b][\/latex] can be partitioned. taking [latex]a=a_0<a_1< \\cdots < a_N=b[\/latex] such that the average value of [latex]f[\/latex] over each subinterval [latex][a_{i-1},a_i]=1[\/latex] is equal to 1 for each [latex]i=1\\, \\cdots , N[\/latex]. Explain why the average value of [latex]f[\/latex] over [latex][a,b][\/latex] is also equal to 1.<\/p>\r\n\r\n

    \r\n\r\n[reveal-answer q=\"fs-id1170572370327\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572370327\"]\r\n

    The integral of [latex]f[\/latex] over an interval is the same as the integral of the average of [latex]f[\/latex] over that interval. Thus, [latex]\\begin{array}{l} \\displaystyle\\int_a^b f(t) dt=\\displaystyle\\int_{a_0}^{a_1} f(t) dt + \\displaystyle\\int_{a_1}^{a_2} f(t) dt + \\cdots + \\displaystyle\\int_{a_{N-1}}^{a_N} f(t) dt = \\displaystyle\\int_{a_0}^{a_1} 1 dt + \\displaystyle\\int_{a_1}^{a_2} 1 dt + \\cdots + \\displaystyle\\int_{a_{N-1}}^{a_N} 1 dt \\\\ =(a_1-a_0)+(a_2-a_1)+ \\cdots +(a_N-a_{N-1})=a_N-a_0=b-a. \\end{array}[\/latex]<\/p>\r\nDividing through by [latex]b-a[\/latex] gives the desired identity.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

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    75.\u00a0<\/strong>Suppose that for each [latex]i[\/latex] such that [latex]1\\le i\\le N[\/latex] one has [latex]\\displaystyle\\int_{i-1}^i f(t) dt=i[\/latex]. Show that [latex]\\displaystyle\\int_0^N f(t) dt=\\frac{N(N+1)}{2}[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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    76.\u00a0<\/strong>Suppose that for each [latex]i[\/latex] such that [latex]1\\le i\\le N[\/latex] one has [latex]\\displaystyle\\int_{i-1}^i f(t) dt=i^2[\/latex]. Show that [latex]\\displaystyle\\int_0^N f(t) dt=\\frac{N(N+1)(2N+1)}{6}[\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1170571654044\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571654044\"]\r\n

    [latex]\\displaystyle\\int_0^N f(t) dt=\\underset{i=1}{\\overset{n}{\\Sigma}} \\displaystyle\\int_{i-1}^i f(t) dt=\\underset{i=1}{\\overset{n}{\\Sigma}} i^2=\\frac{N(N+1)(2N+1)}{6}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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    77. [T]<\/strong> Compute the left and right Riemann sums [latex]L_{10}[\/latex]\u00a0and [latex]R_{10}[\/latex]\u00a0and their average [latex]\\frac{L_{10}+R_{10}}{2}[\/latex] for [latex]f(t)=t^2[\/latex] over [latex][0,1][\/latex]. Given that [latex]\\displaystyle\\int_0^1 t^2 dt=0.\\bar{33}[\/latex], to how many decimal places is [latex]\\frac{L_{10}+R_{10}}{2}[\/latex] accurate?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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    78. [T]<\/strong> Compute the left and right Riemann sums, [latex]L_{10}[\/latex]\u00a0and [latex]R_{10}[\/latex], and their average [latex]\\frac{L_{10}+R_{10}}{2}[\/latex] for [latex]f(t)=(4-t^2)[\/latex] over [latex][1,2][\/latex]. Given that [latex]\\displaystyle\\int_1^2 (4-t^2) dt=1.\\bar{66}[\/latex], to how many decimal places is [latex]\\frac{L_{10}+R_{10}}{2}[\/latex] accurate?<\/p>\r\n[reveal-answer q=\"fs-id1170571543369\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571543369\"]\r\n

    [latex]L_{10}=1.815, \\, R_{10}=1.515, \\, \\frac{L_{10}+R_{10}}{2}=1.665[\/latex], so the estimate is accurate to two decimal places.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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    79.\u00a0<\/strong>If [latex]\\displaystyle\\int_1^5 \\sqrt{1+t^4} dt=41.7133 \\cdots[\/latex], what is [latex]\\displaystyle\\int_1^5 \\sqrt{1+u^4} du[\/latex]?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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    80. <\/strong>Estimate [latex]\\displaystyle\\int_0^1 [latex]t[\/latex] dt[\/latex] using the left and right endpoint sums, each with a single rectangle. How does the average of these left and right endpoint sums compare with the actual value [latex]\\displaystyle\\int_0^1 [latex]t[\/latex] dt[\/latex]?<\/p>\r\n[reveal-answer q=\"fs-id1170572367517\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572367517\"]\r\n

    The average is [latex]1\/2[\/latex], which is equal to the integral in this case.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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    81.\u00a0<\/strong>Estimate [latex]\\displaystyle\\int_0^1 [latex]t[\/latex] dt[\/latex] by comparison with the area of a single rectangle with height equal to the value of [latex]t[\/latex] at the midpoint [latex]t=\\frac{1}{2}[\/latex]. How does this midpoint estimate compare with the actual value [latex]\\displaystyle\\int_0^1 [latex]t[\/latex] dt[\/latex]?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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    82.\u00a0<\/strong>From the graph of [latex]\\sin (2\\pi x)[\/latex] shown:<\/p>\r\n\r\n

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    1. Explain why [latex]\\displaystyle\\int_0^1 \\sin (2\\pi t) dt=0[\/latex].<\/li>\r\n \t
    2. Explain why, in general, [latex]\\displaystyle\\int_a^{a+1} \\sin (2\\pi t) dt=0[\/latex] for any value of [latex]a[\/latex].\r\n\"A<\/span><\/li>\r\n<\/ol>\r\n[reveal-answer q=\"fs-id1170571779594\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571779594\"]\r\n

      a. The graph is antisymmetric with respect to [latex]t=\\frac{1}{2}[\/latex] over [latex][0,1][\/latex], so the average value is zero. b. For any value of [latex]a[\/latex], the graph between [latex][a,a+1][\/latex] is a shift of the graph over [latex][0,1][\/latex], so the net areas above and below the axis do not change and the average remains zero.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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      83.\u00a0<\/strong>If [latex]f[\/latex] is 1-periodic [latex](f(t+1)=f(t))[\/latex], odd, and integrable over [latex][0,1][\/latex], is it always true that [latex]\\displaystyle\\int_0^1 f(t) dt=0[\/latex]?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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      84.\u00a0<\/strong>If [latex]f[\/latex] is 1-periodic and [latex]\\displaystyle\\int_0^1 f(t) dt=A[\/latex], is it necessarily true that [latex]\\displaystyle\\int_a^{1+a} f(t) dt=A[\/latex] for all [latex]A[\/latex]?<\/p>\r\n[reveal-answer q=\"fs-id1170572398848\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170572398848\"]\r\n

      Yes, the integral over any interval of length 1 is the same.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>","rendered":"

      In the following exercises, express the limits as integrals.<\/p>\n

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      1.\u00a0<\/strong>[latex]\\underset{n\\to \\infty }{\\lim}\\underset{i=1}{\\overset{n}{\\Sigma}}(x_i^*) \\Delta x[\/latex] over [latex][1,3][\/latex]<\/p>\n<\/div>\n<\/div>\n

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      2.\u00a0<\/strong>[latex]\\underset{n\\to \\infty }{\\lim}\\underset{i=1}{\\overset{n}{\\Sigma}}(5(x_i^*)^2-3(x_i^*)^3) \\Delta x[\/latex] over [latex][0,2][\/latex]<\/p>\n

      Show Solution<\/span><\/p>\n
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      [latex]\\displaystyle\\int_0^2 (5x^2-3x^3) dx[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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      3.\u00a0<\/strong>[latex]\\underset{n\\to \\infty }{\\lim}\\underset{i=1}{\\overset{n}{\\Sigma}} \\sin^2 (2\\pi x_i^*) \\Delta x[\/latex] over [latex][0,1][\/latex]<\/p>\n<\/div>\n<\/div>\n

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      4.\u00a0<\/strong>[latex]\\underset{n\\to \\infty }{\\lim}\\underset{i=1}{\\overset{n}{\\Sigma}} \\cos^2 (2\\pi x_i^*) \\Delta x[\/latex] over [latex][0,1][\/latex]<\/p>\n

      Show Solution<\/span><\/p>\n
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      [latex]\\displaystyle\\int_0^1 \\cos^2 (2\\pi x) dx[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

      In the following exercises, given [latex]L_n[\/latex]\u00a0or [latex]R_n[\/latex]\u00a0as indicated, express their limits as [latex]n\\to \\infty[\/latex] as definite integrals, identifying the correct intervals.<\/p>\n

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      5.\u00a0<\/strong>[latex]L_n=\\frac{1}{n}\\underset{i=1}{\\overset{n}{\\Sigma}}\\frac{i-1}{n}[\/latex]<\/p>\n<\/div>\n<\/div>\n

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      6.\u00a0<\/strong>[latex]R_n=\\frac{1}{n}\\underset{i=1}{\\overset{n}{\\Sigma}}\\frac{i}{n}[\/latex]<\/p>\n

      Show Solution<\/span><\/p>\n
      [latex]\\displaystyle\\int_0^1 x dx[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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      7.\u00a0<\/strong>[latex]L_n=\\frac{2}{n}\\underset{i=1}{\\overset{n}{\\Sigma}}(1+2\\frac{i-1}{n})[\/latex]<\/p>\n<\/div>\n<\/div>\n

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      8.\u00a0<\/strong>[latex]R_n=\\frac{3}{n}\\underset{i=1}{\\overset{n}{\\Sigma}}(3+3\\frac{i}{n})[\/latex]<\/p>\n

      Show Solution<\/span><\/p>\n
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      [latex]\\displaystyle\\int_3^6 x dx[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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      9.\u00a0<\/strong>[latex]L_n=\\frac{2\\pi }{n}\\underset{i=1}{\\overset{n}{\\Sigma}}2\\pi \\frac{i-1}{n} \\cos (2\\pi \\frac{i-1}{n})[\/latex]<\/p>\n<\/div>\n<\/div>\n

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      10.\u00a0<\/strong>[latex]R_n=\\frac{1}{n}\\underset{i=1}{\\overset{n}{\\Sigma}}(1+\\frac{i}{n})\\log((1+\\frac{i}{n})^2)[\/latex]<\/p>\n

      Show Solution<\/span><\/p>\n
      [latex]\\displaystyle\\int_1^2 x\\log(x^2) dx[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n

      In the following exercises (11-16), evaluate the integrals of the functions graphed using the formulas for areas of triangles and circles, and subtracting the areas below the [latex]x[\/latex]-axis.<\/p>\n

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      11.\u00a0<\/strong>\"A<\/span><\/div>\n<\/div>\n
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      12.\u00a0<\/strong>\"A<\/span><\/p>\n

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      Show Solution<\/span><\/p>\n
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      [latex]1+2 \\cdot 2+3 \\cdot 3=14[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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      13.\u00a0<\/strong>\"A<\/span><\/div>\n<\/div>\n
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      14.\u00a0<\/strong>\"A<\/span><\/p>\n
      Show Solution<\/span><\/p>\n
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      [latex]1-4+9=6[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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      15.\u00a0<\/strong>\"A<\/span><\/div>\n<\/div>\n
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      16.\u00a0<\/strong>\"A<\/span><\/p>\n

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      Show Solution<\/span><\/p>\n
      \n

      [latex]1-2\\pi +9=10-2\\pi[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

      In the following exercises (17-24), evaluate the integral using area formulas.<\/p>\n

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      17.\u00a0<\/strong>[latex]\\displaystyle\\int_0^3 (3-x) dx[\/latex]<\/p>\n<\/div>\n<\/div>\n

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      18.\u00a0<\/strong>[latex]\\displaystyle\\int_2^3 (3-x) dx[\/latex]<\/p>\n

      Show Solution<\/span><\/p>\n
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      The integral is the area of the triangle, [latex]\\frac{1}{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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      19.\u00a0<\/strong>[latex]\\displaystyle\\int_{-3}^3 (3-|x|) dx[\/latex]<\/p>\n<\/div>\n<\/div>\n

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      20.\u00a0<\/strong>[latex]\\displaystyle\\int_0^6 (3-|x-3|) dx[\/latex]<\/p>\n

      Show Solution<\/span><\/p>\n
      \n

      The integral is the area of the triangle, 9.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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      21.\u00a0<\/strong>[latex]\\displaystyle\\int_{-2}^2 \\sqrt{4-x^2} dx[\/latex]<\/p>\n<\/div>\n<\/div>\n

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      22.\u00a0<\/strong>[latex]\\displaystyle\\int_1^5 \\sqrt{4-(x-3)^2} dx[\/latex]<\/p>\n

      Show Solution<\/span><\/p>\n
      The integral is the area [latex]\\frac{1}{2}\\pi r^2=2\\pi[\/latex].<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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      23.\u00a0<\/strong>[latex]\\displaystyle\\int_0^{12} \\sqrt{36-(x-6)^2} dx[\/latex]<\/p>\n<\/div>\n<\/div>\n

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      24.\u00a0<\/strong>[latex]\\displaystyle\\int_{-2}^3 (3-|x|) dx[\/latex]<\/p>\n

      Show Solution<\/span><\/p>\n
      \n

      The integral is the area of the \u201cbig\u201d triangle minus the \u201cmissing\u201d triangle, [latex]9-\\frac{1}{2}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

      In the following exercises (25-28), use averages of values at the left ([latex]L[\/latex]) and right ([latex]R[\/latex]) endpoints to compute the integrals of the piecewise linear functions with graphs that pass through the given list of points over the indicated intervals.<\/p>\n

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      25.\u00a0<\/strong>[latex]\\{(0,0),(2,1),(4,3),(5,0),(6,0),(8,3)\\}[\/latex] over [latex][0,8][\/latex]<\/p>\n<\/div>\n<\/div>\n

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      26.\u00a0<\/strong>[latex]\\{(0,2),(1,0),(3,5),(5,5),(6,2),(8,0)\\}[\/latex] over [latex][0,8][\/latex]<\/p>\n

      Show Solution<\/span><\/p>\n
      \n

      [latex]L=2+0+10+5+4=21,R=0+10+10+2+0=22, \\, \\frac{L+R}{2}=21.5[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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      27.\u00a0<\/strong>[latex]\\{(-4,-4),(-2,0),(0,-2),(3,3),(4,3)\\}[\/latex] over [latex][-4,4][\/latex]<\/p>\n<\/div>\n<\/div>\n

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      28.\u00a0<\/strong>[latex]\\{(-4,0),(-2,2),(0,0),(1,2),(3,2),(4,0)\\}[\/latex] over [latex][-4,4][\/latex]<\/p>\n

      Show Solution<\/span><\/p>\n
      \n

      [latex]L=0+4+0+4+2=10,R=4+0+2+4+0=10, \\, \\frac{L+R}{2}=10[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

      Suppose that [latex]\\displaystyle\\int_0^4 f(x) dx=5[\/latex] and [latex]\\displaystyle\\int_0^2 f(x) dx=-3[\/latex], and [latex]\\displaystyle\\int_0^4 g(x) dx=-1[\/latex] and [latex]\\displaystyle\\int_0^2 g(x) dx=2[\/latex]. In the following exercises (29-34), compute the integrals.<\/p>\n

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      29.\u00a0<\/strong>[latex]\\displaystyle\\int_0^4 (f(x)+g(x)) dx[\/latex]<\/p>\n<\/div>\n<\/div>\n

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      30.\u00a0<\/strong>[latex]\\displaystyle\\int_2^4 (f(x)+g(x)) dx[\/latex]<\/p>\n

      Show Solution<\/span><\/p>\n
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      [latex]\\displaystyle\\int_2^4 f(x) dx + \\displaystyle\\int_2^4 g(x) dx=8-3=5[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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      31.\u00a0<\/strong>[latex]\\displaystyle\\int_0^2 (f(x)-g(x)) dx[\/latex]<\/p>\n<\/div>\n<\/div>\n

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      32.\u00a0<\/strong>[latex]\\displaystyle\\int_2^4 (f(x)-g(x)) dx[\/latex]<\/p>\n

      Show Solution<\/span><\/p>\n
      [latex]\\displaystyle\\int_2^4 f(x) dx - \\displaystyle\\int_2^4 g(x) dx=8+3=11[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n
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      33.\u00a0<\/strong>[latex]\\displaystyle\\int_0^2 (3f(x)-4g(x)) dx[\/latex]<\/p>\n<\/div>\n<\/div>\n

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      34.\u00a0<\/strong>[latex]\\displaystyle\\int_2^4 (4f(x)-3g(x)) dx[\/latex]<\/p>\n

      Show Solution<\/span><\/p>\n
      \n

      [latex]4 \\displaystyle\\int_2^4 f(x) dx - 3 \\displaystyle\\int_2^4 g(x) dx = 32+9=41[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

      In the following exercises (35-38), use the identity [latex]\\displaystyle\\int_{\u2212A}^A f(x) dx = \\displaystyle\\int_{\u2212A}^0 f(x) dx + \\displaystyle\\int_0^A f(x) dx[\/latex] to compute the integrals.<\/p>\n

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      35.\u00a0<\/strong>[latex]\\displaystyle\\int_{\u2212\\pi}^{\\pi} \\frac{\\sin t}{1+t^2} dt[\/latex] (Hint<\/em>: [latex]\\sin(\u2212t)=\u2212\\sin (t)[\/latex])<\/p>\n<\/div>\n<\/div>\n

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      36. <\/strong>[latex]\\displaystyle\\int_{\u2212\\sqrt{\\pi}}^{\\sqrt{\\pi}} \\frac{t}{1+ \\cos t} dt[\/latex]<\/p>\n

      Show Solution<\/span><\/p>\n
      \n

      The integrand is odd; the integral is zero.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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      37.\u00a0<\/strong>[latex]\\displaystyle\\int_1^3 (2-x) dx[\/latex] (Hint:<\/em> Look at the graph of [latex]f[\/latex].)<\/p>\n<\/div>\n<\/div>\n

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      38.\u00a0<\/strong>[latex]{\\displaystyle\\int }_{2}^{4}{(x-3)}^{3}dx[\/latex] (Hint:<\/em> Look at the graph of [latex]f[\/latex].)<\/p>\n

      Show Solution<\/span><\/p>\n
      \n

      The integrand is antisymmetric with respect to [latex]x=3[\/latex]. The integral is zero.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

      In the following exercises (39-44), given that [latex]\\displaystyle\\int_0^1 x dx = \\frac{1}{2}, \\, \\displaystyle\\int_0^1 x^2 dx = \\frac{1}{3}[\/latex], and [latex]\\displaystyle\\int_0^1 x^3 dx = \\frac{1}{4}[\/latex], compute the integrals.<\/p>\n

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      39.\u00a0<\/strong>[latex]\\displaystyle\\int_0^1 (1+x+x^2+x^3) dx[\/latex]<\/p>\n<\/div>\n<\/div>\n

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      40.\u00a0<\/strong>[latex]\\displaystyle\\int_0^1 (1-x+x^2-x^3) dx[\/latex]<\/p>\n

      Show Solution<\/span><\/p>\n
      \n

      [latex]1-\\frac{1}{2}+\\frac{1}{3}-\\frac{1}{4}=\\frac{7}{12}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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      41.\u00a0<\/strong>[latex]\\displaystyle\\int_0^1 (1-x)^2 dx[\/latex]<\/p>\n<\/div>\n<\/div>\n

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      42.\u00a0<\/strong>[latex]\\displaystyle\\int_0^1 (1-2x)^3 dx[\/latex]<\/p>\n

      Show Solution<\/span><\/p>\n
      \n

      [latex]\\displaystyle\\int_0^1 (1-6x+12x^2-8x^3) dx = (x-3x^{2}+4x^{3}-2x^{4})=(1-3+4-2)=0[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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      43.\u00a0<\/strong>[latex]\\displaystyle\\int_0^1 (6x-\\frac{4}{3}x^2) dx[\/latex]<\/p>\n<\/div>\n<\/div>\n

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      44.\u00a0<\/strong>[latex]\\displaystyle\\int_0^1 (7-5x^3) dx[\/latex]<\/p>\n

      Show Solution<\/span><\/p>\n
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      [latex]7-\\frac{5}{4}=\\frac{23}{4}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

      In the following exercises (45-50), use the comparison theorem.<\/p>\n

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      45.\u00a0<\/strong>Show that [latex]\\displaystyle\\int_0^3 (x^2-6x+9) dx \\ge 0[\/latex].<\/p>\n<\/div>\n<\/div>\n

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      46.\u00a0<\/strong>Show that [latex]\\displaystyle\\int_{-2}^3 (x-3)(x+2) dx \\le 0[\/latex].<\/p>\n

      Show Solution<\/span><\/p>\n
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      The integrand is negative over [latex][-2,3][\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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      47.\u00a0<\/strong>Show that [latex]\\displaystyle\\int_0^1 \\sqrt{1+x^3} dx \\le \\displaystyle\\int_0^1 \\sqrt{1+x^2} dx[\/latex].<\/p>\n<\/div>\n<\/div>\n

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      48.\u00a0<\/strong>Show that [latex]\\displaystyle\\int_1^2 \\sqrt{1+x} dx \\le \\displaystyle\\int_1^2 \\sqrt{1+x^2} dx[\/latex].<\/p>\n

      Show Solution<\/span><\/p>\n
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      [latex]x \\le x^2[\/latex] over [latex][1,2][\/latex], so [latex]\\sqrt{1+x} \\le \\sqrt{1+x^2}[\/latex] over [latex][1,2][\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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      49.\u00a0<\/strong>Show that [latex]\\displaystyle\\int_0^{\\pi\/2} \\sin [latex]t[\/latex] dt \\ge \\frac{\\pi}{4}[\/latex]. (Hint<\/em>: [latex]\\sin [latex]t[\/latex] \\ge \\frac{2t}{\\pi}[\/latex] over [latex][0,\\frac{\\pi}{2}][\/latex])<\/p>\n<\/div>\n<\/div>\n

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      50.\u00a0<\/strong>Show that [latex]\\displaystyle\\int_{\u2212\\pi\/4}^{\\pi\/4} \\cos [latex]t[\/latex] dt \\ge \\pi \\sqrt{2}\/4[\/latex].<\/p>\n

      Show Solution<\/span><\/p>\n
      \n

      [latex]\\cos (t) \\ge \\frac{\\sqrt{2}}{2}[\/latex]. Multiply by the length of the interval to get the inequality.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

      In the following exercises (51-56), find the average value [latex]f_{\\text{ave}}[\/latex]\u00a0of [latex]f[\/latex] between [latex]a[\/latex] and [latex]b[\/latex], and find a point [latex]c[\/latex], where [latex]f(c)=f_{\\text{ave}}[\/latex].<\/p>\n

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      51.\u00a0<\/strong>[latex]f(x)=x^2, \\, a=-1, \\, b=1[\/latex]<\/p>\n<\/div>\n<\/div>\n

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      52.\u00a0<\/strong>[latex]f(x)=x^5, \\, a=-1, \\, b=1[\/latex]<\/p>\n

      Show Solution<\/span><\/p>\n
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      [latex]f_{\\text{ave}}=0; \\, c=0[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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      53.\u00a0<\/strong>[latex]f(x)=\\sqrt{4-x^2}, \\, a=0, \\, b=2[\/latex]<\/p>\n<\/div>\n<\/div>\n

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      54.\u00a0<\/strong>[latex]f(x)=(3-|x|), \\, a=-3, \\, b=3[\/latex]<\/p>\n

      Show Solution<\/span><\/p>\n
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      [latex]\\frac{3}{2}[\/latex] when [latex]c= \\pm \\frac{3}{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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      55.\u00a0<\/strong>[latex]f(x)= \\sin x, \\, a=0, \\, b=2\\pi[\/latex]<\/p>\n<\/div>\n<\/div>\n

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      56.\u00a0<\/strong>[latex]f(x)= \\cos x, \\, a=0, \\, b=2\\pi[\/latex]<\/p>\n

      Show Solution<\/span><\/p>\n
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      [latex]f_{\\text{ave}}=0; \\, c=\\frac{\\pi}{2},\\frac{3\\pi}{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

      In the following exercises, approximate the average value using Riemann sums [latex]L_{100}[\/latex]\u00a0and [latex]R_{100}[\/latex]. How does your answer compare with the exact given answer?<\/p>\n

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      57.\u00a0[T]<\/strong> [latex]y=\\ln (x)[\/latex] over the interval [latex][1,4][\/latex]; the exact solution is [latex]\\dfrac{\\ln (256)}{3}-1[\/latex].<\/p>\n<\/div>\n<\/div>\n

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      58. [T]\u00a0<\/strong>[latex]y=e^{x\/2}[\/latex] over the interval [latex][0,1][\/latex]; the exact solution is [latex]2(\\sqrt{e}-1)[\/latex].<\/p>\n

      Show Solution<\/span><\/p>\n
      \n

      [latex]L_{100}=1.294, \\, R_{100}=1.301[\/latex]; the exact average is between these values.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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      59. [T]\u00a0<\/strong>[latex]y= \\tan x[\/latex] over the interval [latex][0,\\frac{\\pi}{4}][\/latex]; the exact solution is [latex]\\dfrac{2\\ln (2)}{\\pi}[\/latex].<\/p>\n<\/div>\n<\/div>\n

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      60. [T]\u00a0<\/strong>[latex]y=\\dfrac{x+1}{\\sqrt{4-x^2}}[\/latex] over the interval [latex][-1,1][\/latex]; the exact solution is [latex]\\frac{\\pi }{6}[\/latex].<\/p>\n

      Show Solution<\/span><\/p>\n
      \n

      [latex]L_{100} \\times (\\frac{1}{2})=0.5178, \\, R_{100} \\times (\\frac{1}{2})=0.5294[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

      In the following exercises, compute the average value using the left Riemann sums [latex]L_N[\/latex]\u00a0for [latex]N=1,10,100[\/latex]. How does the accuracy compare with the given exact value?<\/p>\n

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      61. [T]\u00a0<\/strong>[latex]y=x^2-4[\/latex] over the interval [latex][0,2][\/latex]; the exact solution is [latex]-\\frac{8}{3}[\/latex].<\/p>\n<\/div>\n<\/div>\n

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      62. [T]\u00a0<\/strong>[latex]y=xe^{x^2}[\/latex] over the interval [latex][0,2][\/latex]; the exact solution is [latex]\\frac{1}{4}(e^4-1)[\/latex].<\/p>\n

      Show Solution<\/span><\/p>\n
      \n

      [latex]L_1=0, \\, L_{10} \\times (\\frac{1}{2})=8.743493, \\, L_{100} \\times (\\frac{1}{2})=12.861728[\/latex]. The exact answer [latex]\\approx 26.799[\/latex], so [latex]L_{100}[\/latex]\u00a0is not accurate.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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      63. [T]\u00a0<\/strong>[latex]y=\\left(\\frac{1}{2}\\right)^x[\/latex] over the interval [latex][0,4][\/latex]; the exact solution is [latex]\\dfrac{15}{64\\ln (2)}[\/latex].<\/p>\n<\/div>\n<\/div>\n

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      64. [T]\u00a0<\/strong>[latex]y=x \\sin (x^2)[\/latex] over the interval [latex][\u2212\\pi ,0][\/latex]; the exact solution is [latex]\\dfrac{\\cos (\\pi^2)-1}{2\\pi}[\/latex].<\/p>\n

      Show Solution<\/span><\/p>\n
      \n

      [latex]L_1 \\times (\\frac{1}{\\pi})=1.352, \\, L_{10} \\times (\\frac{1}{\\pi})=-0.1837, \\, L_{100} \\times (\\frac{1}{\\pi})=-0.2956[\/latex]. The exact answer [latex]\\approx -0.303[\/latex], so [latex]L_{100}[\/latex]\u00a0is not accurate to first decimal.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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      65.\u00a0<\/strong>Suppose that [latex]A=\\displaystyle\\int_0^{2\\pi} \\sin^2 [latex]t[\/latex] dt[\/latex] and [latex]B=\\displaystyle\\int_0^{2\\pi} \\cos^2 [latex]t[\/latex] dt[\/latex]. Show that [latex]A+B=2\\pi[\/latex] and [latex]A=B[\/latex].<\/p>\n<\/div>\n<\/div>\n

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      66.\u00a0<\/strong>Suppose that [latex]A=\\displaystyle\\int_{\u2212\\pi\/4}^{\\pi\/4} \\sec^2 [latex]t[\/latex] dt = \\pi [\/latex] and [latex]B=\\displaystyle\\int_{\u2212\\pi\/4}^{\\pi\/4} \\tan^2 [latex]t[\/latex] dt[\/latex]. Show that [latex]B-A=\\frac{\\pi }{2}[\/latex].<\/p>\n

      Show Solution<\/span><\/p>\n
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      Use [latex]\\tan^2 \\theta +1= \\sec^2 \\theta[\/latex]. Then, [latex]B-A=\\displaystyle\\int_{\u2212\\pi\/4}^{\\pi\/4} 1 dx = \\frac{\\pi}{2}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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      67.\u00a0<\/strong>Show that the average value of [latex]\\sin^2 t[\/latex] over [latex][0,2\\pi][\/latex] is equal to [latex]\\frac{1}{2}[\/latex]. Without further calculation, determine whether the average value of [latex]\\sin^2 t[\/latex] over [latex][0,\\pi][\/latex] is also equal to [latex]\\frac{1}{2}[\/latex].<\/p>\n<\/div>\n<\/div>\n

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      68.\u00a0<\/strong>Show that the average value of [latex]\\cos^2 t[\/latex] over [latex][0,2\\pi][\/latex] is equal to [latex]1\/2[\/latex]. Without further calculation, determine whether the average value of [latex]\\cos^2 (t)[\/latex] over [latex][0,\\pi][\/latex] is also equal to [latex]1\/2[\/latex].<\/p>\n

      Show Solution<\/span><\/p>\n
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      [latex]\\displaystyle\\int_0^{2\\pi} \\cos^2 [latex]t[\/latex] dt = \\pi[\/latex], so divide by the length [latex]2\\pi[\/latex] of the interval. [latex]\\cos^2 t[\/latex] has period [latex]\\pi[\/latex], so yes, it is true.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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      69.\u00a0<\/strong>Explain why the graphs of a quadratic function (parabola) [latex]p(x)[\/latex] and a linear function [latex]\\ell (x)[\/latex] can intersect in at most two points. Suppose that [latex]p(a)=\\ell (a)[\/latex] and [latex]p(b)=\\ell (b)[\/latex], and that [latex]\\displaystyle\\int_a^b p(t) dt > \\displaystyle\\int_a^b \\ell (t) dt[\/latex]. Explain why [latex]\\displaystyle\\int_c^d p(t) > \\displaystyle\\int_c^d \\ell (t) dt[\/latex] whenever [latex]a \\le c < d \\le b[\/latex].<\/p>\n<\/div>\n<\/div>\n

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      70.\u00a0<\/strong>Suppose that parabola [latex]p(x)=ax^2+bx+c[\/latex] opens downward [latex](a<0)[\/latex] and has a vertex of [latex]y=\\frac{\u2212b}{2a}>0[\/latex]. For which interval [latex][A,B][\/latex] is [latex]\\displaystyle\\int_A^B (ax^2+bx+c) dx[\/latex] as large as possible?<\/p>\n

      Show Solution<\/span><\/p>\n
      \n

      The integral is maximized when one uses the largest interval on which [latex]p[\/latex] is nonnegative. Thus, [latex]A=\\frac{\u2212b-\\sqrt{b^2-4ac}}{2a}[\/latex] and [latex]B=\\frac{\u2212b+\\sqrt{b^2-4ac}}{2a}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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      71.\u00a0<\/strong>Suppose [latex][a,b][\/latex] can be subdivided into subintervals [latex]a=a_0\n

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      1. Explain why [latex]\\displaystyle\\int_a^b f(t) dt = A_1+A_2+ \\cdots +A_N[\/latex].<\/li>\n
      2. Then, explain why [latex]|\\displaystyle\\int_a^b f(t) dt| \\le \\displaystyle\\int_a^b |f(t)| dt[\/latex].<\/li>\n<\/ol>\n<\/div>\n<\/div>\n
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        72.\u00a0<\/strong>Suppose [latex]f[\/latex] and [latex]g[\/latex] are continuous functions such that [latex]\\displaystyle\\int_c^d f(t) dt \\le \\displaystyle\\int_c^d g(t) dt[\/latex] for every subinterval [latex][c,d][\/latex] of [latex][a,b][\/latex]. Explain why [latex]f(x)\\le g(x)[\/latex] for all values of [latex]x[\/latex].<\/p>\n

        Show Solution<\/span><\/p>\n
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        If [latex]f(t_0)>g(t_0)[\/latex] for some [latex]t_0 \\in [a,b][\/latex], then since [latex]f-g[\/latex] is continuous, there is an interval containing [latex]t_0[\/latex] such that [latex]f(t)>g(t)[\/latex] over the interval [latex][c,d][\/latex], and then [latex]\\displaystyle\\int_c^d f(t) dt>\\displaystyle\\int_c^d g(t) dt[\/latex] over this interval.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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        73.\u00a0<\/strong>Suppose the average value of [latex]f[\/latex] over [latex][a,b][\/latex] is 1 and the average value of [latex]f[\/latex] over [latex][b,c][\/latex] is 1 where [latex]a\n<\/div>\n<\/div>\n

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        74.\u00a0<\/strong>Suppose that [latex][a,b][\/latex] can be partitioned. taking [latex]a=a_0\n

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        Show Solution<\/span><\/p>\n
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        The integral of [latex]f[\/latex] over an interval is the same as the integral of the average of [latex]f[\/latex] over that interval. Thus, [latex]\\begin{array}{l} \\displaystyle\\int_a^b f(t) dt=\\displaystyle\\int_{a_0}^{a_1} f(t) dt + \\displaystyle\\int_{a_1}^{a_2} f(t) dt + \\cdots + \\displaystyle\\int_{a_{N-1}}^{a_N} f(t) dt = \\displaystyle\\int_{a_0}^{a_1} 1 dt + \\displaystyle\\int_{a_1}^{a_2} 1 dt + \\cdots + \\displaystyle\\int_{a_{N-1}}^{a_N} 1 dt \\\\ =(a_1-a_0)+(a_2-a_1)+ \\cdots +(a_N-a_{N-1})=a_N-a_0=b-a. \\end{array}[\/latex]<\/p>\n

        Dividing through by [latex]b-a[\/latex] gives the desired identity.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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        75.\u00a0<\/strong>Suppose that for each [latex]i[\/latex] such that [latex]1\\le i\\le N[\/latex] one has [latex]\\displaystyle\\int_{i-1}^i f(t) dt=i[\/latex]. Show that [latex]\\displaystyle\\int_0^N f(t) dt=\\frac{N(N+1)}{2}[\/latex].<\/p>\n<\/div>\n<\/div>\n

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        76.\u00a0<\/strong>Suppose that for each [latex]i[\/latex] such that [latex]1\\le i\\le N[\/latex] one has [latex]\\displaystyle\\int_{i-1}^i f(t) dt=i^2[\/latex]. Show that [latex]\\displaystyle\\int_0^N f(t) dt=\\frac{N(N+1)(2N+1)}{6}[\/latex].<\/p>\n

        Show Solution<\/span><\/p>\n
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        [latex]\\displaystyle\\int_0^N f(t) dt=\\underset{i=1}{\\overset{n}{\\Sigma}} \\displaystyle\\int_{i-1}^i f(t) dt=\\underset{i=1}{\\overset{n}{\\Sigma}} i^2=\\frac{N(N+1)(2N+1)}{6}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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        77. [T]<\/strong> Compute the left and right Riemann sums [latex]L_{10}[\/latex]\u00a0and [latex]R_{10}[\/latex]\u00a0and their average [latex]\\frac{L_{10}+R_{10}}{2}[\/latex] for [latex]f(t)=t^2[\/latex] over [latex][0,1][\/latex]. Given that [latex]\\displaystyle\\int_0^1 t^2 dt=0.\\bar{33}[\/latex], to how many decimal places is [latex]\\frac{L_{10}+R_{10}}{2}[\/latex] accurate?<\/p>\n<\/div>\n<\/div>\n

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        78. [T]<\/strong> Compute the left and right Riemann sums, [latex]L_{10}[\/latex]\u00a0and [latex]R_{10}[\/latex], and their average [latex]\\frac{L_{10}+R_{10}}{2}[\/latex] for [latex]f(t)=(4-t^2)[\/latex] over [latex][1,2][\/latex]. Given that [latex]\\displaystyle\\int_1^2 (4-t^2) dt=1.\\bar{66}[\/latex], to how many decimal places is [latex]\\frac{L_{10}+R_{10}}{2}[\/latex] accurate?<\/p>\n

        Show Solution<\/span><\/p>\n
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        [latex]L_{10}=1.815, \\, R_{10}=1.515, \\, \\frac{L_{10}+R_{10}}{2}=1.665[\/latex], so the estimate is accurate to two decimal places.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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        79.\u00a0<\/strong>If [latex]\\displaystyle\\int_1^5 \\sqrt{1+t^4} dt=41.7133 \\cdots[\/latex], what is [latex]\\displaystyle\\int_1^5 \\sqrt{1+u^4} du[\/latex]?<\/p>\n<\/div>\n<\/div>\n

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        80. <\/strong>Estimate [latex]\\displaystyle\\int_0^1 [latex]t[\/latex] dt[\/latex] using the left and right endpoint sums, each with a single rectangle. How does the average of these left and right endpoint sums compare with the actual value [latex]\\displaystyle\\int_0^1 [latex]t[\/latex] dt[\/latex]?<\/p>\n

        Show Solution<\/span><\/p>\n
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        The average is [latex]1\/2[\/latex], which is equal to the integral in this case.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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        81.\u00a0<\/strong>Estimate [latex]\\displaystyle\\int_0^1 [latex]t[\/latex] dt[\/latex] by comparison with the area of a single rectangle with height equal to the value of [latex]t[\/latex] at the midpoint [latex]t=\\frac{1}{2}[\/latex]. How does this midpoint estimate compare with the actual value [latex]\\displaystyle\\int_0^1 [latex]t[\/latex] dt[\/latex]?<\/p>\n<\/div>\n<\/div>\n

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        82.\u00a0<\/strong>From the graph of [latex]\\sin (2\\pi x)[\/latex] shown:<\/p>\n

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        1. Explain why [latex]\\displaystyle\\int_0^1 \\sin (2\\pi t) dt=0[\/latex].<\/li>\n
        2. Explain why, in general, [latex]\\displaystyle\\int_a^{a+1} \\sin (2\\pi t) dt=0[\/latex] for any value of [latex]a[\/latex].
          \n\"A<\/span><\/li>\n<\/ol>\n
          Show Solution<\/span><\/p>\n
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          a. The graph is antisymmetric with respect to [latex]t=\\frac{1}{2}[\/latex] over [latex][0,1][\/latex], so the average value is zero. b. For any value of [latex]a[\/latex], the graph between [latex][a,a+1][\/latex] is a shift of the graph over [latex][0,1][\/latex], so the net areas above and below the axis do not change and the average remains zero.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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          83.\u00a0<\/strong>If [latex]f[\/latex] is 1-periodic [latex](f(t+1)=f(t))[\/latex], odd, and integrable over [latex][0,1][\/latex], is it always true that [latex]\\displaystyle\\int_0^1 f(t) dt=0[\/latex]?<\/p>\n<\/div>\n<\/div>\n

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          84.\u00a0<\/strong>If [latex]f[\/latex] is 1-periodic and [latex]\\displaystyle\\int_0^1 f(t) dt=A[\/latex], is it necessarily true that [latex]\\displaystyle\\int_a^{1+a} f(t) dt=A[\/latex] for all [latex]A[\/latex]?<\/p>\n

          Show Solution<\/span><\/p>\n
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          Yes, the integral over any interval of length 1 is the same.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n\n\t\t\t

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